The power factor of a CR circuit is frac{1}{s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The power factor of a $CR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C} = \frac{1}

Vedclass · Accepted Answer

$\frac{1}{\sqrt{5}}$

Vedclass · Answer

(B) The power factor of a $CR$ circuit is given by $\cos \phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.Given $\cos \phi_1 = \frac{1}{\sqrt{2}}$,this implies $\frac{R}{Z} = \frac{1}{\sqrt{2}}$,so $R = X_{C1} = \frac{1}{2\pi f_1 C}$.When the frequency is halved,$f_2 = \frac{f_1}{2}$.The new capacitive reactance is $X_{C2} = \frac{1}{2\pi f_2 C} = \frac{1}{2\pi (f_1/2) C} = 2 X_{C1} = 2R$.The new power factor is $\cos \phi_2 = \frac{R}{\sqrt{R^2 + X_{C2}^2}} = \frac{R}{\sqrt{R^2 + (2R)^2}} = \frac{R}{\sqrt{R^2 + 4R^2}} = \frac{R}{\sqrt{5R^2}} = \frac{1}{\sqrt{5}}$.

The power factor of a $CR$ circuit is $\frac{1}{\sqrt{2}}$. If the frequency of an $AC$ signal is halved,then the power factor of the circuit will become:

Similar Questions

In the circuit shown here,the voltages across $L$ and $C$ are respectively $300\, V$ and $400\, V$. The voltage $E$ of the $AC$ source is......$V$.

In a series $L-C-R$ circuit,

Vedclass Test Series

Exam Paper Generator

Online Exam Module